George has a quadratic of the form $x^2+bx+\frac13$, where $b$ is a specific negative number. Using his knowledge of how to complete the square, George is able to rewrite this quadratic in the form $(x+m)^2+\frac{1}{12}$. What is $b$?
Explanation: The expansion of $(x+m)^2+\frac{1}{12}$ is $x^2+2mx+m^2+\frac{1}{12}$, which has a constant term of $m^2+\frac{1}{12}$. This constant term must be equal to the constant term of the original quadratic, so $$m^2+\frac{1}{12} = \frac13,$$and $$m^2 = \frac13-\frac{1}{12} = \frac14.$$This yields the possibilities $m=\frac12$ and $m=-\frac12$.

If $m=\frac12$, then $(x+m)^2+\frac{1}{12} = x^2+x+\frac14+\frac{1}{12} = x^2+x+\frac13$. This implies $b=1$, but we reject this possibility because we were told that $b$ is a negative number.

If $m=-\frac12$, then $(x+m)^2+\frac{1}{12} = x^2-x+\frac14+\frac{1}{12} = x^2-x+\frac13$, giving the result $b=\boxed{-1}$.